If $\mathrm{n}=24,\overline{x}=34$, and $s=6$, construct a confidence interval at a $98\mathrm{\%}$ confidence level. Assume the data came from a normally distributed population.

Give your answers to one decimal place.
Can we conclude that the population mean is at least 29.04?
No
Yes

Can we conclude that the population mean is no more than 38.56 ?
No
Yes
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Step 1 ：Given: $n=24$, $\overline{x}=34$, $s=6$, and a 98% confidence level.

Step 2 ：The Z-score for a 98% confidence level is approximately 2.33 (you can find this value in a standard Z-table).

Step 3 ：Substitute these values into the formula for the confidence interval: $CI=\overline{x}\pm Z\ast (s/\sqrt{n})$

Step 4 ：Calculate the value inside the parentheses: $6/\sqrt{24}=1.224744871$

Step 5 ：Multiply this by the Z-score: $2.33\ast 1.224744871=2.85365347$

Step 6 ：So, the confidence interval is: $CI=34\pm 2.85365347$

Step 7 ：This gives us a range from: $34-2.85365347=31.14634653\approx 31.1$ (to one decimal place)

Step 8 ：to: $34+2.85365347=36.85365347\approx 36.9$ (to one decimal place)

Step 9 ：So, the 98% confidence interval is $\boxed{{\displaystyle (31.1,36.9)}}$

Step 10 ：Can we conclude that the population mean is at least 29.04? No, because 29.04 is not within our confidence interval.

Step 11 ：Can we conclude that the population mean is no more than 38.56? Yes, because 38.56 is greater than the upper limit of our confidence interval.